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# Find the Taylor series about the given point for each of the functions. In each case find the radius of convergence. $f(x)=\sin x^{2}$ about $x_{0}=0$

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$\textbf{Solution}$: Let us consider an analytic function $f$ defined by

$f(x)=\sin x^2.$

We will now find the power series expansion of the function $f$ about the point $\>x_0=0.$ First consider that $\> y=x^2$. Then our reformed function is $\> f(y) =\sin y.$ And also note that $\> x_0^2=y_0=0.$ Now notice that

\begin{align}f(y)=\sum_{n=0}^{\infty} \frac{f^{(n)}(y_0)}{n!}(y-y_0)^n,\end{align}

for all $y$ in the interval of convergence. Recall that, $\> f(y)=\sin y.$ Now notice that

\begin{align*} & f^{(1)}(y)=\cos y \\ & f^{(2)}(y)=- \sin y \\ & f^{(3)}(y)=-\cos y \\ & f^{(4)}(y)=\sin y \\ & \cdots \cdots \end{align*}

Then notice that, at $y_0=0$ we have

$f^{(2n)}(0)=0 \>\>\> \text{ and }\>\>\> f^{(2n+1)}(0)=(-1)^{n}.$

Then from $(1)$ we have

$f(y)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} y^{2n+1}.$

Then notice that

$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1}.$

Thus the Taylor series expansion of $f(x)=\sin x^2$ is given by

$\sin x^2= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1},$

for all $x$ in the interval of convergence.

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